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y : \R \times \R^3 \rightarrow \R^2
as the
solution of our ode forward problem
.
Suppose we are also given measurement values
z_k \in \R
for
k = 1, 2, 3, 4
that are modeled by
\[
z_k = y_1 ( s_k , a) + e_k
\]
where
s_k \in \R
,
e_k \sim {\bf N} (0 , \sigma^2 )
,
and
\sigma \in \R_+
.
The maximum likelihood estimate for
a
given
( z_1 , z_2 , z_3 , z_4 )
solves the following inverse problem
\[
\begin{array}{rcl}
{\rm minimize} \;
& \sum_{k=1}^4 H_k [ y( s_k , a ) , a ]
& \;{\rm w.r.t} \; a \in \R^3
\end{array}
\]
where the function
H_k : \R^2 \times \R^3 \rightarrow \R
is
defined by
\[
H_k (y, a) = ( z_k - y_1 )^2
\]
a
one uses a
numerical procedure
to solve for
y_1 ( s_k , a )
for
k = 1 , 2 , 3, 4
.
\[
\begin{array}{rcl}
{\rm minimize}
& \sum_{k=1}^4 H_k( y^{k * ns} , a )
& \; {\rm w.r.t} \; y^1 \in \R^2 , \ldots , y^{4 * ns} \in \R^2 ,
\; a \in \R^3
\\
{\rm subject \; to}
& y^{M+1} = y^M +
\left[ G( y^M , a ) + G( y^{M+1} , a ) \right] * \frac{ s_{M+1} - s_M }{ 2 }
& \; {\rm for} \; M = 0 , \ldots , 4 * ns - 1
\\
& y^0 = F(a)
\end{array}
\]
where
ns
is the number of time intervals
(used by the trapezoidal approximation)
between each of the measurement times.
Note that, in this form, the iterations of the optimization procedure
also solve the forward problem equations.
In addition, the functions that need to be differentiated
do not involve an iterative procedure.